# Find the mass of one element of a sample

• Chemistry
• jackthehat
In summary, we have 2 moles of K in 1 mole of K2CrO4. For 1 mole of substance, we have 78.1966 g. So how many moles of the substance do we have if our sample contains just 0.67 g of K? We have 0.00856 moles. The mass of K2CrO4 in 0.00856 moles is (0.00856 x 194.1903) g.f

#### jackthehat

Homework Statement
A sample of potassium chromate, K2CrO4, is analyzed and found to contain exactly 0.67 grams of potassium. What is the mass of this sample of potassium chromate?
Relevant Equations
Formula mass = (2 x K) + (1 x Cr) + (4 x O)
1 mole of K2CrO4 has formula mass of ((2 x K) + (1 x Cr) + (4 x O)) g
=> ((2 x 39.0983) + (1 x 51.9961) + (4 x 15.9994)) g = 194.1903 g
we have 2 moles of K in 1 mole of K2CrO4 = (2x 39.0983) g = 78.1966 g
so for 1 mole of substance we have 78.1966 g
So how many moles of the substance do we have if our sample contains just 0.67 g of K
=> 1 mole x (0.67/78.1966) = 0.00856 moles
So the mass of K2CrO4 in 0.00856 moles is ( 0.00856 x194.1903) g
=> Mass of K2CrO4 of this sample is 1.66386 g
Now I was not sure of the number of significant figures required for the solution
since we were originally given the amount 0.67 g of K
I was unsure if this is to 2 or 3 significant figures so I gave the answer to both
I put down 1.66 g (to 3 sig fig) and 1.7 g (to 2 sig fig)
But both answers appear to be incorrect … can anyone help I am stuck ?

As far as I can tell, your numbers are right. A more direct way to do it might have been to use the ratio $$\frac{2\times 0.39}{2\times0.39 + 52 + 4\times 16} = \frac{0.67}{M}$$

jim mcnamara
As far as I can tell, your numbers are right. A more direct way to do it might have been to use the ratio $$\frac{2\times 0.39}{2\times0.39 + 52 + 4\times 16} = \frac{0.67}{M}$$
Hi etotheipi,
Thank you for taking the time to look at my problem and for your advice. it has been a help to me to see the problem tackled in an alternative way. much appreciated.
Regards,
jackthehat

BillTre and etotheipi
You meant to write 39 not 0.39?

The correct answer is 1.7, not 1.66 or 1.67. This question it's from coursera!
0.67gK* (1molK/39.098gK)*(1molK2CrO4)*(194.188gK2CrO4/1molK2CrO4) = 1.7g (not 1.66g or 1.67g)

K = 2*39.098
Cr=51.996
O=4*15.999
K2CrO4=194.188 g

The correct answer is 1.7, not 1.66 or 1.67.
That's a matter of significant figures. Yes, data contains only two, so the answer should contain two as well. But significant figures are faulty by design and most people I know are not too religious about them.

Unfortunately some homework checking systems are very strict, which leaves impression as if SF really mattered. They don't. They do help explaining the intuition behind answer like 1.66385112651956734 not making any sense, but that's all.

Besides, OP stated 1.7 wasn't accepted as the correct answer.

I would never use the elements' atomic mass to six significant figures ! - two or three is the normal way to work, as several of the others have used.

While you're doing the right math, I think it's most reassuring if you put everything into a standard format where you can cross out the units. Start by asking yourself what the question is, fill in the unit you need to end up with as part of a conversion factor, and continue until you have your answer:

? g K2CrO4 = (____ g K2CrO4 / ____ mol K2CrO4) x ( ____ mol K2CrO4 / ____ mol K) x (____ mol K / ____ g K) x ( ___ g K)

Then just fill in the numbers so all the conversion factors (the things in parentheses) are equal to 1

= (194.19 g K2CrO4 / 1 mol K2CrO4) x (1 mol K2CrO4 / 2 mol K) x ( 1 mol K / 39.10 g K) x (0.67 g K)
= 1.7 g K2CrO4

Now to be sure, that's what everyone else got, so there's probably a mistranscription somewhere, but at the first line, with all the blanks unfilled, you should know you have the right answer, and the rest is just button-pushing.

On the other hand, maybe it's the sig figs. They say it's a mass of EXACTLY 0.67 g, don't they? As in 0.6700000000...? :) [no, seriously, nobody's that insane. I hope. Until I read a newspaper that is.]